I've attached a sketch of the way I think the forces involved in a rocker suspension work on the chassis. The assumptions are that the sprung weight of the corner is 150kg, and that the pivot on the rocker (point X) is in the middle.

Have I got this right?

What about the lateral (left/right) forces? That top pivot will be hit with a lot of lateral force!

Hi Mr H. My view in my flued up state is the following.

Point Y. As it's a 1:1 ratio 150kgs in equal 150 kgs out so agreed.

Point X. Disagree here. Effectively the cars weight is all sitting on this pivot so it's 150 kg for me.

ETA. Doh.. Of course the rocker is effectively a 2:1 lever thus doubling the force to 300 kg. Matt, you were right and I was wrong.

Cheers,
Mike

[Edited on 3/12/09 by MikeCapon]

I'm no chassis engineer, just a physicist but.


If the rocker is equal length then...

The chassis is pushing up at Y with 150kg force (yuck I know I should use N). In the same way that the rocker at the other end is being pushed up by the wheel.

Therefore the rocker at X is being pulled away from the chassis with a 300kg (3000N force).

I hope that makes some sense. You can think of it by thinking what would happen if you took out a bolt (where would it go).

Matt

quote:

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Originally posted by matt_gsxr

You can think of it by thinking what would happen if you took out a bolt (where would it go).

Matt

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Assuming the corner weight is 150kg taking moments about the inner upper connection tells you with a 1:1 ratio the middle pivot must be twice the 150kg.

Matt is right



[Edited on 3/12/09 by liam.mccaffrey]

[Edited on 3/12/09 by liam.mccaffrey]

I had to work all this out for my design and if I recall (a long time ago!) matt-gsxr is correct. The pivot sees the corner weight plus the reaction from the spring, i.e. corner weight x rocker ratio (only 1 in your 1:1 example) trying to pull it off the chassis. In other words a rocker pivot has a hard life! Even more so if you have something more than a 1:1 ratio on the rocker as I do - i.e. shorter length inboard.

Liam

I'm with matt_gsxr on this one.

Hugh

^^^
What he said

The fulcrum reaction equals the sum of the two end reactions. 150 + 150 = 300 I believe.

Cheers R

Thanks for that chaps, makes very good sense.

and of course these are only static loads
if the inboard section is shorter then the loads are higher (wouldnt it??)

e.g if the inboard section is half the length the leverage is halved, therefore the spring and dampner assembly would be under a 300kg static load and the pivot 450kg static load.. ??

this seems right in my head

aitch

Also be aware of the very high bending moment exerted by the lower end of the spring-damper unit on lower member.

[Edited on 3/12/09 by britishtrident]

This probably explains why most inboard shock absorbers are connected to the suspension via a pushrod and a bellcrank. In Staniforths book I noticed the rocker/wishbone set ups would create these very significant loads and have the disadvantage of being difficult to alter or adjust in some ways.

quote:

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Originally posted by Benonymous
This probably explains why most inboard shock absorbers are connected to the suspension via a pushrod and a bellcrank.

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and more connections linkages and pivots in a bell crank design, but it is more adjustable by its nature and more options to position of the spring dampner assembly

aitch

[Edited on 9/12/09 by aitch]

There's something wrong with this picture; since the car isn't moving, the total forces must total to zero.

There's 150kg pushing the outer end of the rocker upward, and since the rocker's balanced (not rotating), it means there's also 150kg upward force acting on the inboard end. Therefore, there's 150kg downward force on the bottom of the spring, so there's 150kg net upward force on the rocker-arm pivot, not 75kg.



[Edited on 12/9/09 by kb58]

quote:

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Originally posted by kb58
There's something wrong with this picture; since the car isn't moving, the total forces must total to zero.

There's 150kg pushing the outer end of the rocker upward, and since the rocker's balanced (not rotating), it means there's also 150kg upward force acting on the inboard end. Therefore, there's 150kg downward force on the bottom of the spring, so there's 150kg net upward force on the rocker-arm pivot, not 75kg.

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I think the problem, maybe with my own opinion, is what the frame of reference is. I'm trying to look at just the frame - no tire - in order to remove the confusion of an external reference.

I'll revisit this later when I have more time... maybe I need to pull out my statics book...

In a nutshell, you can consider that the 300kg pushing the rocker pivot upwards, is resisted by the 150kg spring force plus the 150kg corner weight pusing downwards.

The way I see it:

quote:

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Originally posted by Badger_McLetcher
The way I see it:

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wow....this is just like being back at college doing my H.N.D.

have you considered a single large damper mounted horizontal?

[Edited on 9/12/09 by boggle]